[image] Semantics of :: in return type vs. argument type annotations There is another potential source of confusion, which is that = returns the value on the right, pre-conversion: julia> x::Float64 = 1 1 julia> x 1.0 This maybe my biggest syntax gripe in julia. I really think that assignmen…

[image] Semantics of :: in return type vs. argument type annotations No, x = y = 1 is the same as x = (y = 1), and 1 is assigned to both variables (the sensible choice) regardless of y::Float64’s conversion, so the subexpression y = 1 must have a value of 1. Again the choice of either side wouldn…

[image] Mason: do you really read that as temp = (;a=1, b=2, c=3, d=4) (;a, b) = temp x = temp Yes, out of habit (Python). I think it’s better this way: even if we don’t let assignment convert, indexing and properties may do more drastic changes, so x = y[1] = 0 resulting in y[1] = 0; x = 0 …

[image] Mason: Regarding this argument, I don’t really find it makes much sense to me. When you see someone write x = (;a, b) = (;a=1, b=2, c=3, d=4) do you really read that as temp = (;a=1, b=2, c=3, d=4) (;a, b) = temp x = temp There isn’t really a correct answer here, because there’s n…

[image] mnemnion: You expect it to be right associative: (x = (y = z)), but it’s left associative ((x = y) = z). Assignment is right-associative, but the value of an assignment expression is the right-hand side, not the left. (x = y) = z is not a valid expression, because x = y is not an l-va…

[image] stevengj: Assignment is right-associative, but the value of an assignment expression is the right-hand side Is there a language where chained assignment is left-associative? AFAIK it’s all right-associative, it’s just a tossup whether the value is the right hand side (which becomes th…

Note that as of 1.11, Array is mostly implemented in Julia (although the new Memory which Array is now built on still has significant amounts of C)

[image] Semantics of :: in return type vs. argument type annotations Assignment is right-associative, but the value of an assignment expression is the right-hand side, not the left. (x = y) = z is not a valid expression, because x = y is not an l-value. Thank you, this is what I was attempting…

I guess it’s just that for me, the only way I can really understand syntax like x::T = y (a, b) = z (;a, b) = w is that these aren’t real assignments, nor are they real pattern matching, but they’re actually transformed-assignments, so it seems weird to me that these transformations don’t get prop…

[image] Mason: is that these aren’t real assignments, nor are they real pattern matching, but they’re actually transformed-assignments The reason assignment isn’t propagated is because assignment applies transformations like conversion and destructuring. That means that the several L-values c…

[image] Benny: Is there a language where chained assignment is left-associative? AFAIK it’s all right-associative, it’s just a tossup whether the value is the right hand side (which becomes the value assigned to every target) or the left hand side (which becomes a chain of possible changes via a…

Why assignment operators return the right-hand-side

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Why assignment operators return the right-hand-side

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